3x^2+42=27x

Solution for 3x^2+42=27x equation:

3x^2+42=27x

We move all terms to the left:

3x^2+42-(27x)=0

a = 3; b = -27; c = +42;
Δ = b2-4ac
Δ = -272-4·3·42
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-15}{2*3}=\frac{12}{6}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+15}{2*3}=\frac{42}{6}
=7
$